Guide :

# what is the largest rectangular area that can be covered with 20 meters of fence??

## Research, Knowledge and Information :

### Fence It : nrich.maths.org

What is the largest rectangular area you can fence off ... 10 by 20 rectangle was the best. Well done. Can you ... the area covered? And what is the length of fence ...

### A farmer has 230 ft of fence to enclose a rectangular garden ...

... what is the largest rectangular area he can ... If he has 20 m of fence , what is the largest area he ... 2. a farmer has 400 meters of fence with which to ...

### Geometry Word Problems: Maximizing and Minimizing

Geometry Word Problems: ... The largest possible area is 1024 square meters ...and I ... "The rectangle with the largest area for a given perimeter will be a square ...

### Word Problems: Quadratic Max/Min Application - Rectangular Areas

... Quadratic Max/Min Application - Rectangular Areas: ... wants to find the largest rectangular area that can be fenced off. ... To help the fencing cover more land, ...

### A rectangular garden with an area of 250 square meters is to ...

Find the dimensions of the largest area he can fence. ... is the area of the part of the garden NOT covered ... are 20 meters long Calculus 1) A rectangular ...

### If you have 40 feet of fencing to put around a garden what ...

... of the largest rectangular garden he can ... than 20 ft, and the width can be ... to fence the largest possible rectangular area using 200 ...

### Farmer Ed has 550 meters of fencing, and wants to enclose a ...

... and wants to enclose a rectangular plot that borders on ... maximize the area. What is the largest area that can be ... rectangular plot be x meters then ...

### Math Quick Take: Optimizing your garden's area - Bob Sutor

Math Quick Take: Optimizing your garden’s area. July 27, ... What’s the largest rectangular area garden you can enclose with it? ... then l = 20 and the area is ...

### Optimization Problem #4 - Max Area Enclosed by Rectangular ...

May 03, 2011 · Optimization Problem #4 - Maximizing the Area of Rectangular Fence Using Calculus ... Maximizing the Area of Rectangular Fence Using Calculus ... 20 ...

## Suggested Questions And Answer :

### what is the largest rectangular area that can be covered with 20 meters of fence??

#### area not "kuvered" , yu  put fens AROUND area best shape=square...1 side^2= 20 meters fens/4=5 meters for eech side area=5^2=25 squares

### my fence or rectangular area will be 20x6

Define the shape of the rectangular area by establishing a relationship between the length and width of the rectangle. For example, L = 2W + 5, or W = 3L – 4. Be sure to include the appropriate units (inches, feet, yards, miles, or meters). Using the fact that A = LW, together with the relationship defined in step 2, eliminate one of the variables to set up a quadratic equation. Solve the quadratic equation using any of the techniques learned in this unit. The solution(s) will be one of the dimensions; use step 2 to find the other. Now determine the perimeter so that you will know how much fencing to buy

### A sheep pen

This answer has now been revised. I hope it's correct according to the given figures. Sorry for the error. I must have confused two very similar questions. The perimeter of the pen is 2L+2W where L and W are length and width. (a) The fence has a length 2L+W because the shed's length is 30m and forms a width. Since W=30, 2L+30=200, 2L=170, L=85m. The area is 30*85=2550 sq m. [corrected] (b) The maximum area for the pen is a square, so L=W and the area is L^2. The perimeter is 4L. We know that the shed helps to form a side. So the perimeter is 200+30=230m=4L, so L=57.5m and the area is 3306.25 sq m. [corrected] It's easy to prove that a square is the maximum area for a given perimeter. If L=a-x and W=a+x, the fixed perimeter, P,  is 4a, so a=P/4. The area is a^2-x^2. This quantity has a maximum value of a^2 only when x=0, which means L=W=a, a square. (c) (i) Now, if the length of each pen is L and the total amount of fencing is 200m, then we can write an equation for the total amount of fencing:  3L+3W=200, where W=30, since the shed covers the width of one pen, so L=110/3=36 2/3m. (ii) If the two pens are square for maximum area each, and the side is L, the total enclosed perimeter is 7L. The perimeter is made up of 200m of fence and 30m of shed=230m. L=230/7=32 6/7m=32.857m approx. [corrected] (d) For three pens of maximum area the perimeter is 10L=200+30=230, and L=23m. The area of each pen is 529 sq m. [corrected] Total pen area is 1587 sq m.

### The owner of rectangular lot wants to put a fence around the area using 120 meter barb wire. The length of the rectangular lot is x. Express the area in square meters of the field as a function of x.

The wire represents the perimeter of the lot=2(length+width)=120, or length+width=60, so, if length=x, x+width=60 and width=60-x metres. The area is length times width=x(60-x)=60x-x^2. We can write f(x)=60x-x^2 where area=f(x) in square metres.

### You have a 500 foot roll of fencing and a large field. You want to construct a rectangular playgorund area. What are the dimensions of the largest such yard? What is the largest area?

The perimeter of the rectangle is 2(L+W) where L and W are length and width. So L+W=250'. W=250-L. Area=L(250-L). This area is maximum when L=125, area=125^2=15625 sq ft. (rectangle is a square). To prove this let L=125+a and W=125-a. Perimeter=2(125+a+125-a)=500. Area=125^2-a^2, which is always <125^2 except when a=0. So the maximum area is 125^2.

### Quadratic Functions, Dividing Polynomials, and Zeros of Polynomials

1. The maximum area is when the rectangle is a square. So the three fenced sides are equal and have length 150/3=50 yd. Let's look at the logic: the sides of the rectangle are L and W so the area, A, is LW and the perimeter, P=2L+2W. L=(P-2W)/2 and A=W(P-2W)/2=(1/2)(WP-2W^2). The maximum value of A is obtained by differentiating the quadratic with respect to W: (1/2)(P-4W)=0 at a turning point, so W=P/4 and L=(P-P/2)/2=P/4. Therefore L=W=P/4. The enclosed area is therefore a square, and we know that the three sides plus another unfenced length, L, make up the perimeter P=4L=150+L, so 3L=150 and L=50 yd. Alternative solution avoiding calculus: a. A=LW where L=length of rectangular area, W=width. b. 150=2L+W. c. W=150-2L. But the true perimeter of the area is P=2L+2W, so W=(1/2)(P-2L). d. A=L(P-2L)/2=(LP-2L^2)/2=LP/2-L^2. e. f(L)=-L^2+LP/2-A in standard quadratic form f(x)=ax^2+bx+c, where a=-1, b=P/2, c=-A and x=L. f. -b/2a=(-P/2)/(-2)=P/4. This is the vertex of f(L), and L=P/4 is the value for the maximum area. g. From (c) W=(1/2)(P-2L)=P/4, so W=L=P/4. This makes the area a square of area P^2/16. From (b) 150=2L+W=P/2+P/4=3P/4, making P=4/3*150=200, so W=L=200/4=50 yd. h. Maximum area is 50^2=2500 sq yd. 2. Synthetic division (1) -3 | 2 -3 -45 -54 ......2 -6..27...54 ......2 -9 -18 | 0 3. Synthetic division (2) 6 | 2..-9 -18 .....2 12..18 .....2...3..| 0 This quotient is 2x+3, representing the third factor of the polynomial. 4. Synthetic division (3) -4 | 2 -3 -45..-54 ......2 -8..44.....4 ......2 -11 -1 | -50 The Remainder Theorem tells us that substituting x=-4 into the polynomial gives us the same remainder. The substitution gives: -128-48+180-54=-50. The zeroes are -3, 6 and -3/2, because synthetic division (2) tells us that 2x+3 is a factor.

### Find length and cost ?

Let length of land be x. Thus, perimeter of land = 50m + 50m + x + x = 100m + 2x Cost of fencing land is Rs 18 per meter, so cost of fencing land will be = (100 + 2x) * 18 = 1800 + 36x The actual cost is Rs 4680, so we have: 4680 = 1800 + 36x 36x = 4680 - 1800 36x = 2880 x = 2880 / 36 x = 80 Thus, length of land is 80m. Area of land = length * width = 80m * 50m = 4000m^2 Cost of leveling plot = Rs7.6 * 4000 = Rs 30400

a. A=LW where L=length of rectangular area, W=width. b. 150=2L+W. c. W=150-2L. But the true perimeter of the area is P=2L+2W, so W=(1/2)(P-2L). d. A=L(P-2L)/2=LP/2-L^2. e. f(L)=-L^2+LP/2-A. f. -b/2a=(-P/2)/(-2)=P/4. This is the vertex of f(L), and L=P/4 is the value for the maximum area. g. From (c) W=(1/2)(P-2L)=P/4, so W=L=P/4. This makes the area a square of area P^2/16. From (b) 150=2L+W=P/2+P/4=3P/4, making P=4/3*150=200, so W=L=200/4=50 yd. h. Maximum area is 50^2=2500 sq yd.

### Find the cost ?

lenght of room =10m breadth of room=7.5m Area of room=10*7.5=75m cost of carpet per meter=250 cost of carpet per 75 meters=250*75=18750.

### rectangular area question?

x= one of the 3 parallel segments 2000-3x left to divide equally for the other two sides x/2(2000-3x) =-3/2 x^2 +1000x this is an inverted parabola and the vertex of the graph will give the maximum x. -b/2a =-1000 / 2(-3/2) =1000/3 yards for each of the three parallel sides leaving 500 yards each for the other two sides 500(1000/3) =500000/3 sq yds is the max area